Divide the following complex numbers. $\dfrac{-23+11i}{5+i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${5-i}$. $ \dfrac{-23+11i}{5+i} = \dfrac{-23+11i}{5+i} \cdot \dfrac{{5-i}}{{5-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-23+11i) \cdot (5-i)} {5^2 - (i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-23+11i) \cdot (5-i)} {(5)^2 - (i)^2} $ $ = \dfrac{(-23+11i) \cdot (5-i)} {25 + 1} $ $ = \dfrac{(-23+11i) \cdot (5-i)} {26} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-23+11i}) \cdot ({5-i})} {26} $ $ = \dfrac{{-23} \cdot {5} + {11} \cdot {5 i} + {-23} \cdot {-1 i} + {11} \cdot {-1 i^2}} {26} $ $ = \dfrac{-115 + 55i + 23i - 11 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{-115 + 55i + 23i + 11} {26} = \dfrac{-104 + 78i} {26} = -4+3i $